Nhanh nha mình cần gấp 

ghi cả cách trình bày ra nhé

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

ĐỀ LỘN XỘN Z. 1/N(N+1) CHỨ

ai nhanh mk k

trả lời em dùm câu hỏi phía dưới vs

Câu a:

A = 1.2 + 2.3 + 3.4 + ...+ 99.100

3A = 1.2.3 + 2.3.3 +..+ 99.100.3

1.2.3 = 1.2.3

2.3.3 = 2.3(4 - 1) = 2.3.4 - 1.2.3

3.4.4 = 3.4(5 - 2) = 3.4.5 - 2.3.4

.............................................................

99.100.3 = 99.100.(101 - 98)=99.100.101-98.99.100

Cộng vế với vế ta có:

3A = 99.100.101

A = 99.100.101 : 3

A = 333300

Câu b:

1.3 + 3.5 + 5.7 + ..+ 97.99

6.A = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 97.99.6

1.3.6 = 1.3.(5 + 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5.(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7

.......................................................

97.99.6 = 97.99.(101 - 95) = 97.99.101-95.97.99

Cộng vế với vế ta có:

6A = 1.3.1 + 97.99.101

A = (1.3.1 + 97.99.101) : 6

A = (3 + 9603.101) : 6

A = (3+ 969903) : 6

A = 969906 : 6

A = 161651

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

Câu C sai đề rùi!

Đề đúng : C =1/1.2 + 1/2.3 +......+ 1/49.50

C = 1 - 1/2 + 1/2 - 1/3 +......+1/49 - 1/50

C = 1 - 1/50

C = 49/50

D = 2/3.5 + 2/5.7 +.......+ 2/37.39

D = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/37 - 1/39

D = 1/3 - 1/39

D = 12/39

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(M=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(M=2\left(1-\frac{1}{100}\right)\)

\(M=2.\frac{99}{100}\)

\(M=\frac{99}{50}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)

\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\frac{98}{99}\)

\(N=\frac{49}{33}\)